Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1

X = 10 is fine but it doesn't help. We know that this inequality holds for all x. We need to plug in a value for x which tells us something about A. If we put x = 0, we are left with just A and that will tell us something about A. Just plugging in any value may not work; you have to look for a smart value.
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So for this question, no other value of x(except x=0,2) gives us any information about A. hence we take only that value which gives us explicit and fixed result for A. Am I correct??
Similarly we would do in other similar questions as well...??

No, look, we know that \(x^2 - 2x + A > 0\) for all x. For every value of x, this inequality should be satisfied.

Put x = 0, you get A > 0
Put x = 1, you get \(1 - 2 + A > 0 i.e. A > 1\)
Put x = 10, you get A > -80

Now the point is that A should take a value such that all these conditions are satisfied. Say A can be 5. If A is 5, it is > 0, > 1 and > -80.

When I look at \(x^2 - 2x + A > 0\) given x can take any value, the first value that pops in my head to get a sense of A is x = 0. That provides exactly what I need. Had the question been whether A > 2, x = 0 would not have helped. I would have had to search a little for a pattern to see how the value of x changes. This question is made in a way that x = 0 helps immediately. Anyway, it is a good idea to try the value 0 in many circumstances. It simplifies things immensely and usually helps you eliminate a couple of options at least.
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if we put x=-3 in 1, then A can have -1 or -2 value also ??

No, A can never have a value of -1 or -2. It must be greater than 1.
If we put x = -3, we get
X^2-2X+A > 0
A > -15
So this tells us that A must be greater than -15. Putting other values of x such as 0, 1, 2 etc tell us that A must be greater than 0 and A must be greater than 1 etc. Since this inequality holds for ALL values of x, A must be greater than 1 because a value greater than 1 will automatically be greater than -15 as well as 0. If we take a value of A such as -14, it will be greater than -15 but not greater than 0 or 1 hence the inequality will not hold for ALL value of x.
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Is A positive?

1) x^2 -2x + A is positive for all x
2) Ax^2 + 1 is positive for all x

My answer is E = Both together are not sufficient. However, test answer say A - Statement 1 is sufficient.

Explanation for statement 1 not sufficient

Case 1: A>0 say 6 then equation becomes x^2 - 2x + 6. Now if x=6 then equation will be equal to 30 which is greater than 0
Case 2: A 0.

Can someone please explain if I am missing something here or the answer is incorrect.

Thanks
Akshay

You haven't used a crucial piece of information in Statement 1:

x^2 -2x + A is positive for all x

That is, it must be positive for every value of x, not only for x = 6. In particular, x^2 - 2x + A must be positive when x = 0, so 0^2 - 2*0 + A > 0, or A > 0, and the statement is sufficient.

Statement 2 is not sufficient because A could be positive, or could be zero.
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Yes..

For statement 1, simply plug in x=0 to find that A>0.

Ian, in statement 2, A can be positive, 0 or negative (Say x=1/2 and A=-1/2).

Also, Akshay, you cannot assume a value for A first and then go about proving that the inequality is positive. It says that the inequality is positive for all values of x NOT for all values of A.

Hope this helps.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is A positive?

(1) x^2-2x+A is positive for all x
(2) Ax^2+1 is positive for all x

There is one variable (A) and 2 equations are given by the conditions, so there is high chance (D) will become the answer.
For condition 1, from y=ax^2+bx+c, if D(discriminant)=b^2-4ac<0, y>0 works for all a>0.
From y=x^2-2x+A, D=(-2)^2-4*1*A<0, 4<4A, 1<A, which answers the question 'yes', makes the condition sufficient.
condition 2 answers the question 'yes' for A=1, but 'no' for A=0, making the condition insufficient.
The answer therefore becomes (A).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

1)\(x^2-2x+A\) is positive for all x
This becomes x^2-2x+A>0
Most if the above solutions talk of Quadratic equation but equation are of form ax^2-2x+A=0 and it is easy to get confused in discriminant etc..
So a simplified way of looking at this..
\(x^2-2x+A>0\) for all values of x....
So least value of (x^2-2x)+A>0

So what is least value ...
It is at x=1, x^2-2x=1^2-1*2=-1
So -1+A>0...A>1
Hence sufficient..
2) Ax^2+1>0
Ax^2>-1...
x^2 is always positive for all values so A can take both negative and positive values..
Insufficient

A
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No, if Statement 2 is true, A cannot be negative. Statement 2 tells us that: Ax^2 + 1 is positive for all x, and not only for x = 1/2. In your example, where A = -1/2, you can quickly see that Ax^2 + 1 will be negative for x = 10, for example, so A cannot be -1/2; that disagrees with the information in the statement. You can see that, if A is negative, Ax^2 + 1 will be zero (so certainly not positive) whenever \(x = \frac{1}{\sqrt{ |A| }\). Plugging that in for x:

\(Ax^2 + 1 = A \left( \frac{1}{\sqrt{ |A| } \right)^2 + 1 = \frac{A}{|A|} + 1 = -1 + 1 = 0\)

and if \(|x| > \frac{1}{\sqrt{ |A| }\), then Ax^2 + 1 will be negative, if A is negative. So if Statement 2 is true, A cannot be negative. A can, however, be zero, which is why the statement is insufficient.
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Statement I is suff, huh? I don't want to sound like I'm degrading anyone's fancy math - but there seems to be a serious disconnect happening here.

I - x^2-2X+A has to be positive for ALL xs. Okay - so all I have to do is find at least 1 x value where A can be either positive or negative.

x=-1 ---> 1+2+A=positive ---> 3+A=positive. My math may not be as fancy as everyone else's here but I'm pretty sure A needs to be greater than negative 3 with this X value thus A has the potential to be a positive or negative value.

A should not be suff...?

I believe this is a GMAT Club question created by Bunnel. You can literally take a semester's worth of a course to study the theory on a single one of the questions Bunnel creates. I don't think most people can gasp the genius thought on some of these questions created by Bunnel, IanStewart and a few others.

Obviously the confusion here is what the heck "all x" means.
This question by no means is wrong. It is however very poorly written at least as far as GMAT standards are concerned. The question is very confusing and not even remotely in the ballpark of what I'd consider fair game for the GMAT.

WRONG
If you are getting Statement (1) wrong it is because you are assuming "all x" means for any specific given value of x. If you do that then you'll start picking numbers to plug into the equation x^2 - 2x + a > 0
For example if x = 5, then
x^2 - 2x + a > 0
(5)^2 - 2(5) + a > 0
25-10+a>0
15+a>0
a>-15

Here you can see a>-15 and say a can be -14 or a can be 30 and say insufficient. But, this is wrong because "all x" does not mean any specific value of x.

CORRECT
"all x" means EVERY single possible value of x. This basically means x is EVERY number in the universe. X is 5, X is 0, X is the amount of money in my checking account (-56.20) , X is be the weight of the earth (5.972 × 10^24). The question means for EVERY value of x, x^2 - 2x + a > 0.
If you simply the right part of the equation you get:
x(x-2) + a > 0
To simply it even more, we can assign a value of "b" to the left part of the equation, so x(x-2) = b. So now we have b + a > 0

Again, the question asks for EVERY value of x, is a in the above equation positive?
Let's try WHEN b =0 (noticed I said WHEN b = 0, not "if" b=0. There isn't an if b=0, it is when b=0)
so for b= x(x-2) =0, then x = -2 or 0. plug x=0 into x(x-2) + a > 0. 0 + a > 0, so a > 0.
Let's try WHEN b = negative number, so b= x(x-2) < 0, then x is between 0 and 2, so let's try x = 1 into the equation x(x-2) + a > 0. -1 + a>0, so a>1.
Let's try WHEN b = positive number, so b= x(x-2) > 0, then x is less than 0 -or- x is more than 2. Let's try x=3 into the equation x(x-2) + a > 0. So 3+a>0, then a > -3.

So, the solution set so far is a>0, a>1, a>-3, we can try different numbers to get more of a solution set, but there is no need at this point, since the most restrictive of the solution set is a>1, so we know a is at least >1, so a is positive.

Why in S2 i cannot use D<0?

it gives me b^2-4ac<0
here b is 0,
so 0-4A<0;
A>0.